Question

A vertical pole (more than $100m$ high) consists of two portions, the lower being one-third of the whole. If the upper portion subtends an angle ${\tan }^{-1}\frac{1}{2}$ at a point in a horizontal plane through the foot of the pole and distance 40 ft from it, then the height of the pole is:

• A. 100 ft
• B. 120 ft
• C. 150 ft
• D. none of these

Answer 1

Answer: (B)

Let the height of the pole $BC$ be $h$. In $△ABC$,
$\tan \beta =\frac{h}{40}$ ..... (i)

and in $\Delta \alpha \beta D,$
$\tan \alpha =\frac{h/3}{40}$
$=\frac{h}{120}$... (ii)
Now, $\tan \theta =\frac{1}{2}(given)$
$\Rightarrow$ $\tan (\beta -\alpha )=\frac{1}{2}$
$\Rightarrow$ $\frac{\tan \beta -\tan \alpha }{1+\tan \beta \tan \alpha }=\frac{1}{2}$
$\Rightarrow$ $\frac{\frac{3h}{120}-\frac{h}{120}}{1+\frac{3h^2}{14400}}=\frac{1}{2}$
$\Rightarrow$ $h=120,40$ But $h$
cannot be taken according to the given condition, therefore $h=120ft.$